A drum of radius R and mass M, rolls down without slipping along an inclined plane of angle `theta`. The frictional force-

Options

A. Decreases the rotational and translational motion
B. Dissipates energy as heat
C. Decreases the rotational motion
D. Converts translational energy to rotational energy

Answer

To solve the problem of a drum of radius \( R \) and mass \( M \) rolling down an inclined plane of angle \( \theta \) without slipping, we need to analyze the forces acting on the drum and apply Newton's second law along with the rotational dynamics.

### Step-by-Step Solution:

1. **Identify the Forces:**
The forces acting on the drum are:
- Gravitational force \( \vec{F_g} = M \vec{g} \) acting vertically downwards.
- Normal force \( \vec{N} \) acting perpendicular to the inclined plane.
- Frictional force \( \vec{F_f} \) acting up the incline.

2. **Resolve the Gravitational Force:**
The gravitational force can be resolved into two components:
- Parallel to the incline: \( F_{\text{parallel}} = Mg \sin \theta \)
- Perpendicular to the incline: \( F_{\text{perpendicular}} = Mg \cos \theta \)

3. **Apply Newton's Second Law:**
For the translational motion down the incline, we can apply Newton's second law:
\[ Mg \sin \theta - F_f = Ma \]
where \( a \) is the linear acceleration of the drum.

4. **Relate Linear and Angular Acceleration:**
Since the drum rolls without slipping, the linear acceleration \( a \) is related to the angular acceleration \( \alpha \) by:
\[ a = R \alpha \]

5. **Torque Equation:**
The frictional force also creates a torque about the center of the drum:
\[ \tau = F_f R = I \alpha \]
For a solid cylinder, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} M R^2 \]
Substituting \( \alpha = \frac{a}{R} \):
\[ F_f R = \left(\frac{1}{2} M R^2\right) \left(\frac{a}{R}\right) \]
Simplifying gives:
\[ F_f = \frac{1}{2} M a \]

6. **Substitute \( F_f \) into the First Equation:**
Now we substitute \( F_f \) back into the equation from step 3:
\[ Mg \sin \theta - \frac{1}{2} M a = Ma \]
Rearranging gives:
\[ Mg \sin \theta = \frac{3}{2} Ma \]
Therefore, we can solve for \( a \):
\[ a = \frac{2}{3} g \sin \theta \]

7. **Calculate the Frictional Force:**
Now substituting \( a \) back into the equation for \( F_f \):
\[ F_f = \frac{1}{2} M a = \frac{1}{2} M \left(\frac{2}{3} g \sin \theta\right) = \frac{1}{3} Mg \sin \theta \]

### Final Result:
The frictional force acting on the drum as it rolls down the incline is:
\[ F_f = \frac{1}{3} Mg \sin \theta \]